https://file.aiquickdraw.com/mathbot/file/1729035920035bgwpljc6.jpg

The problem asks for the values of \(x\) at which the graph of the function:

\[
f(x) = \frac{2x^3}{3} - 4x^2 + 8
\]

has a horizontal tangent. To solve this, we need to find when the derivative \(f'(x)\) is equal to zero, since a horizontal tangent occurs where the slope of the function is zero.

### Step 1: Differentiate the function
Let's first calculate the derivative of \(f(x)\):

\[
f(x) = \frac{2x^3}{3} - 4x^2 + 8
\]

Using basic differentiation rules:

\[
f'(x) = \frac{d}{dx} \left( \frac{2x^3}{3} \right) - \frac{d}{dx} \left( 4x^2 \right)
\]

\[
f'(x) = 2x^2 - 8x
\]

### Step 2: Set the derivative equal to zero
To find when the tangent is horizontal, we solve:

\[
2x^2 - 8x = 0
\]

Factor the equation:

\[
2x(x - 4) = 0
\]

This gives us two solutions:

\[
x = 0 \quad \text{or} \quad x = 4
\]

### Step 3: Conclusion
The function has horizontal tangents at \(x = 0\) and \(x = 4\). Therefore, the correct answer is:

\[
\boxed{\text{Para } x = 0 \text{ y } x = 4}
\]

This corresponds to the second option in the given choices.

---

Would you like more details on this solution, or do you have any questions?

Here are 5 related questions to deepen your understanding:
1. What is the geometric interpretation of a horizontal tangent line?
2. How do we identify critical points of a function using its derivative?
3. What does the second derivative tell us about the concavity of a function at these points?
4. How does the behavior of the function change at \(x = 0\) and \(x = 4\)?
5. Can a function have more than one horizontal tangent in the same interval?

**Tip**: Always remember that horizontal tangents occur where the derivative equals zero.

¿Para qué valores de x, la gráfica de f(x) = (2x^3)/3 - 4x^2 + 8 tiene una tangente horizontal?

Learn how to find the values of x where the function f(x) = (2x^3)/3 - 4x^2 + 8 has horizontal tangents by calculating its derivative and solving f'(x) = 0. This problem involves key calculus concepts suitable for Grades 11-12.

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